Using the equation:
2Al + 2KOH + 4H2SO4 + 22H2O --> 2K[Al(SO4)2] . 12H2O + 3H2
how many grams of alum can be obtained from 20.0 g of aluminum when the reaction proceeds with 100% yield? How many grams of alum would be obtained if the reaction were to proceed with 80.0% yield?
